Can N be constructed?
in [Logic]
|
Prev: "NO BOX CONTAINS THE BOX NUMBERS THAT DON'T CONTAIN THEIR OWN BOX NUMBER" ~ XEN
Next: "NO BOX CONTAINS THE BOX NUMBERS THAT DON'T CONTAIN THEIR OWN BOX NUMBER" ~ XEN
From: WM on 11 Jun 2010 01:38 On 10 Jun., 14:39, jbriggs444 <jbriggs...(a)gmail.com> wrote: > > Re-cast > the problem in terms of limits Here it is: Limits and Dedekind cuts exist by finite definitions only. Their number remains always countable (even finite). Therefore uncountability is nonsense also from this point of view. Regards, WM
From: WM on 11 Jun 2010 01:41 On 10 Jun., 21:44, Virgil <Vir...(a)home.esc> wrote: > > That would be correct, unless a removal is not executed before the > > next line has been established. > > > Therefore the set of remaining lines cannot be empty. > > Then which lines are left?- My proof does not show which line is left. But it shows that finished infinity is a non-mathematical notion. Of course it is always the last line that is left, and it is impossible to get rid of a last line, though the contents of the last line may change as often as desired. Regards, WM
From: WM on 11 Jun 2010 01:48 On 10 Jun., 22:03, Virgil <Vir...(a)home.esc> wrote: > > That is just the case when constructing N or completeing Cantor's > > list. Without completing the list, the diagonal argument is invalid. > > Cantor merely says that no completed list of binary sequences can > contain all binary sequences, which is obviously true. He says first of all that an infinite list cna be completed. That is matheology. > > Thus if the SET of all completed binary sequences exists at all, It exists in the brains of some cranks and crackpots. That's all. > it > cannot be arranged into a sequence of such sequences. This assertion would be correct, but it is not Cantor's and not that of his followers. > > This so upsets WM that he reacts by denying existence of the set of all > infinite binary sequences. > > But in some axiom systems such a set must exist. Those axiom system are self-contradictory. Just in order to show that, I had the discussion with WH. We see as the resut that he must accept an unacceptable premise, i.e., the dispersion of nodes 1 in the list 1 11 111 .... over more than one line. That can be proved to be wrong, by induction and by construction. > > WM is quite free to avoid such axiom systems himself but has neither the > right nor the power to deny access to them for others. > > > Of course that is nonsense, but it is assumed to be possible by set > > theorists. > > When WM equates the reality of an axiom system with physical reality, he > becomes foolish. Mathematics is physics (V. A. Arnold). Regards, WM
From: WM on 11 Jun 2010 01:50 On 10 Jun., 22:06, Virgil <Vir...(a)home.esc> wrote: > In article > <bc98ca3a-1b0a-439e-b4c6-8a4a0ec1b...(a)w12g2000yqj.googlegroups.com>, > > > > > > WM <mueck...(a)rz.fh-augsburg.de> wrote: > > On 10 Jun., 16:13, William Hughes <wpihug...(a)hotmail.com> wrote: > > > > After any finite number of steps the set of remaining lines > > > cannot be empty. > > > No. After any possible step the set of remaining lines cannot be > > empty. > > > > Look! Over There! A Pink Elephant! > > > > After an infinite number of steps the set of remaining lines > > > cannot be empty. > > > How would you get to an infinite number of steps when each step has > > another finite number? > > There is nothing in the relevant axiom system which requires accessing > infinite cases only by such step-by-step operations on finite cases. > > In fact one such infinite case is built into those axioms The infinite list 1 11 111 .... need not be built step by step. Nevertheless the proof stands that there are not two (or more) lines which are necessary to contain all 1's of the list. Regards, WM
From: WM on 11 Jun 2010 01:53
On 11 Jun., 01:14, "K_h" <KHol...(a)SX729.com> wrote: > "WM" <mueck...(a)rz.fh-augsburg.de> wrote in message > > news:172aadb8-7837-4ae0-86fb-f4cecef342c0(a)w12g2000yqj.googlegroups.com... > On 9 Jun., 21:54, William Hughes <wpihug...(a)hotmail.com> > wrote: > > > On Jun 9, 4:16 pm, WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > In fact I write nothing, but only construct. > > For insance I construct by division > > 1/9 = 0.111... > > > That is a construction. And that > > construction can be done as > > 0.1 > > 0.11 > > 0.111 > > ... > > Each number in the list is different from 1/9 by a certain > amount, (1/9)/(10^K) for K = 1,2,3,..., and so the Kth > number in the list, f(K), is given by : > > f ( K ) = 1/9 - (1/9)/(10^K) > > Since each natural number K is finite, the term (1/9)/(10^K) > is never zero. Thus, 1/9 appears nowhere in the list. In > the limiting case as K-->oo we get f(K)-->1/9. Since > ALEPH_0 is infinitely larger than any natural number, after > you have constructed the entire list you are no longer in > the list. That is, because of its hugeness, ALEPH_0 > constructions take you to a place, called 1/9, that is > beyond the list. It is not useful to argue verbally. Show two 1's existing in the list and not existing in one line together. You cannot do that. Therefore aleph_0 is not so large, but only a wrong concept that prevents to apply complete induction. Therefore one of the notions must be dropped. I maintain induction! Regards, WM |